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2 years ago

A survey in a magazine that asks readers to return the survey would result in:

Physics

1 answer:

Angelina_Jolie [31]2 years ago

4 0

D - Most likely. Those who read the magazine can choose whether or not to return the survey.

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A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic

xxTIMURxx [149]

Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2$

=2.5m

on the axis of x at x = 2.5

$r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}$

r = 2.5m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T$

(Along z direction)

B)r = 5.00m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2$

=5.00m

on the axis of x at x = 5.0

$r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}$

r = 5.00m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T$

(Along x direction)

7 0

3 years ago

What happens to the energy of a photon if the wavelength is doubled?.

Vladimir [108]

Answer:

Halving the wavelength doubles the frequency and, thus, doubles the energy of the incident photons. This doubles the energy given to each electron, nearly doubling its kinetic energy after it is free from the metal.

8 0

2 years ago

1 point

Oduvanchick [21]

The distance it traveled is 147

5 0

2 years ago

24. A 50-turn coil has a diameter of 15 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.50 T so that

finlep [7]

Answer:

0.1 = 4.42v

1.0 =0.44v

60 = 7.4 × 10^-3 V

Explanation:

Apply the formula of:

Emf = (No of turns × magnetic flux)÷(the given time)

Where your magnetic flux = (0.50 × cross sectional area(pi × 0.075^2))

4 0

3 years ago

A point source is fixed 1.0 m away from a large screen. Call the line normal to the screen surface and passing through the cente

lubasha [3.4K]

Answer:

d=0.02m

Explanation:

The key of the exercise is to be able to graph the problem to be able to develop it. After that you simply proceed through a triangular relationship.

Then I leave the graph of the problem and give the solution.

So things, under the given relationship, we have,

$\frac {0.04} {0.5} = \frac {d} {0.25}$

<em> *The value of 0.04 is taken because the triangle given in the graph must be divided in half, to obtain rectangular triangles.</em>

$d = \frac {0.04 * 0.25} {0.5}$

$d = 0.02m$

3 0

3 years ago